of 1 49
The Moon As A Message And Primal Causes
By
Ian Beardsley
Copyright © 2023
of 2 49
Contents
Radius and Charge of a Proton……………………….3
Introduction………………………………………………………9
0.0 The Mystery of the Moon……………………………….10
1.0 The Mystery of the Moon Taken Further………… 11
2.0 The Messenger……………………………………………..14
3.0 Logos………………………………………………………….19
4.0 Six-Fold Symmetry and The Periodic Table…….26
5.0 Rigorous Formulation of Proton-Seconds………32
6.0 Theory of Compounds………………………………….35
Appendix 1………………………………………………………..41
Appendix 2……………………………………………………….42
Appendix 3……………………………………………………….45
Appendix 4……………………………………………………….47
of 3 49
Radius and Charge of a Proton
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 0.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 0.2
Which describes mass per meter over time, which is:
Equation 0.3
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 0.4
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 0.5
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 0.6
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
α
1
α
2
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s/m
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
of 4 49
We take the square root to get meters:
Equation 0.7
We multiply that with the value we have in equation 0.4:
Equation 0.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 0.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 0.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 0.11
Equation 0.12
If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of
the elements in the periodic table occur for dividing by something less that 1-second. It seems
the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that (See Appendix 1):
Equation 0.13
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive
the second in terms of something else. For now we have the mass of a proton as:
Equation 0.14
I then developed the concept of proton-seconds where time is associated with atoms through the
number of protons they have.
S
p
= 2.953E − 15m
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
h
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6proton s = 6m
p
m
p
=
1E − 26kg ⋅ s
6secon ds
m
c
=
1E − 26kg ⋅ s
1secon d
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
m
p
=
3r
p
18α
2
4πh
Gc
of 5 49
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a
measure of energy over time where space and time are concerned it must play a role. Of course
the radius of a proton plays a role since squared and multiplied by it is the surface area of our
proton embedded in space. The gravitational constant is force produced per kilogram over a
distance, thus it is a measure of how the surrounding space has an effect on the proton giving it
inertia. The speed of light c has to play a role because it is the velocity at which events are
separated through time. The mass of a proton has to play a role because it is a measurement of
inertia itself. And alas the fine structure constant describes the degree to which these factors
have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional
analysis.
Equation 0.15
That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:
Equation 0.16
4π
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
Fig. 1
of 6 49
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements.
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving (Program in Appendix 4):
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry.
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
1
α
2
m
p
h 4π r
2
p
Gc
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
of 7 49
Equation 0.17
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
Fig. 2
of 8 49
Letting v= from our equation for k (page 12 is about k)
And we have
Equation 0.18
We get
Equation 0.19
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6proton s
of 9 49
Introduction
The moon perfectly eclipses the sun as seen from the earth. This is an incredible fact because not
only does the moon have to be the right distance from the sun to do this and the right size, the
sun also has to be the right size. If this is for a reason, perhaps for the reason to let us know we
are here for a reason by forces unknown, then there should be more clues. Perhaps our distant
ancestors from which we evolved when sitting around a fire at night flaking stones into
spearpoints noticed this and it spoke to them. I find if we consider the universe as the unfolding
of six-fold symmetry that we can predict the radius of a proton and its charge as a consequence
of the geometry of space as well as predict the hydrocarbons, the skeletons of biological life.
While my theory for space and time predicts the radius and charge of a proton, we will mainly
consider here the basic equation in its pertinence to the moon, because it suggests a mystery
surrounding it. The purpose of this paper is to explore that mystery of the moon. We propose the
messenger (that behind the lunar size and orbital radius) based it on the proton because the Sun
has to be the size it is, and the earth-moon system the distance it is from the Sun to be in the
habitable zone and we show that the earth/moon/sun system is related to the proton mass and
size. A proposal for a starting point for developing a logos, or universal language, is given by
starting the undertaking with the planetary orbital velocities and it is suggested that the periodic
table is outside the basic mathematical operations for a reason as well a non-chemical, but
rather mathematical, description of the elements and their compounds is initiated.
of 10 49
0.0 The Mystery of the Moon
If in the planet that has life orbiting a star there is an indication to its intelligence that there is a
mystery before it, it is that for our star the Sun, the moon perfectly eclipses it as seen from the
Earth. This is because:
Which are approximately equal. As well we can look at it as:
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver (Ag)
by molar mass, these elements being used for religious and ceremonial jewelry, is the same 1.8:
The sun is gold in color, the moon is silver in color.
(lu n ar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lu n ar − ra diu s)
sol ar − r a dius
=
1,738.1
696,00
= 0.0025
(lu n ar − ra diu s)
(lu n ar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
sol ar − r a dius
ear th − orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
of 11 49
1.0 The Mystery of the Moon Taken Further
We arrived at an equation that suggests the hydrocarbons, which are the skeletons of life
chemistry, are the sixfold unfolding of space and time (See Appendix 4):
One second gives carbon:
Equation 1.1
Six seconds gives 1 proton is hydrogen:
Equation 1.2.
Where alpha is the fine structure constant is the potential energy of an electron in the first
circular orbit divided by its rest mass times the speed of light squared, in other words the
ground state in a hydrogen atom is:
And is the proton radius, the proton mass, G the universal constant of gravitation, c the
speed of light, and h is Planck’s constant. Equations 1 and 2 are masses divided by proton mass
giving protons, but they can cancel leaving seconds. Equations 1 and 2 suggest the second is a
natural unit, and as we will see being at the basis of life and connected to the moon as some kind
of mystery, that there is perhaps something deep and far reaching going on here. If the second is
natural, then it should be in the earth/moon/sun orbital mechanics because the second comes
from the calendar and the calendar comes from the earth/moon/sun orbital mechanics. I find it
is natural:
Equation 1.3.
Where KE means kinetic energy (See Appendix 1). But this refers to planets and moons, can we
derive the second otherwise? I find we can:
Equation 1.4.
Where is the radius of a hydrogen atom, and is Avogadro’s number. (See
Appendix 2 after reading the next page about k).
We used Chandrasekhar’s Limit for the upper limit of the mass of a white dwarf star such that it
won’t collapse
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
α
2
=
U
e
m
e
c
2
r
p
m
p
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
R
H
N
A
𝔼 = 6E 23
of 12 49
Equation 1.5.
Which allows us to determine an intermediary mass between a proton and a star from the
geometric mean between and , where is the mass of a proton :
Equation 1.6.
Equation 1.7.
Warren Giordano noticed that
Equation 1.8.
Without the right units. I noticed since Avogadro’s number is that I
could introduce an equation of state for the periodic tables of the elements:
Equation 1.9.
Which works for any element , is a variable that is Avogadro’s number multiplied by the
number of protons in . The we formulate as an equation of state for the periodic table of
the elements and is Avogadro’s number 6E23. . From this we have a
constant k that pertains to the macrocosmos and microcosmos:
Equation 1.10.
Equation 1.11.
The thing about k is that it is
Equation 1.12.
Where is the earth orbital velocity is 29,790m/s.
This results in from equation 1.1 and 1.3:
Equation 1.13
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
m
p
M
m
p
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
h(1 + α) ⋅ 10
23
= G
6.02 × 10
23
≈ 6 × 10
23
ℍ = 1
gr a m
atom
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
𝔼
N
A
𝔼
N
A
𝔼
N
A
ℍ = 6E 23atom s /gr a m
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
k =
1
773.5
s
m
k v
e
= 6
v
e
k v
e
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
=
1
α
4
m
2
p
⋅
h 4π r
2
p
Gc
of 13 49
Consider the factor
It is actually equal to 1.2 seconds squared, but we can say 1.2 seconds is a rough sketch for the
idea that it is one second, which you can do because in any physical theory there is room for play
for the physics to still serve its function. So we will evaluate it at one second. We have
On the right we have
This is an accuracy of about 95%. Putting the constants on the right and Earth/Sun/Moon on
the left, and substituting our expression for k, we have:
Equation 1.14.
You can never be two careful formulating such things. I show the work in Appendix 3 by hand
verifying the units work and that the expression on the left is approximately the expression on
the right. Here we have the earth/moon/sun system on the left, and the natural constants on
the right. The interesting thing that happens here is that:
Are the earth orbital velocities on both sides of the equation because t equals one second. That is
we have
Equation 1.15.
Equation 1.16.
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
k v
e
= 38.5
1
α
4
m
2
p
⋅
h 4π r
2
p
Gc
= 36.555
v
e
K E
2
moon
K E
2
earth
⋅ Ear th Da y
2
=
3
4
r
2
p
m
3
p
⋅
1
α
4
2c
π
⋅
h
3
G
3
⋅
1
N
A
𝔼
vt
2
= (vt)t = x t = dista n ce × t im e
v
e
K E
2
moon
K E
2
earth
⋅ Ear th Da y
2
= v
e
(1secon d )
2
3
4
r
2
p
m
3
p
⋅
1
α
4
2c
π
⋅
h
3
G
3
⋅
1
N
A
𝔼
= v
e
(1secon d )
2
of 14 49
2.0 The Messenger
Since the message from forces unknown is the moon, then we should take its size as a natural
unit and set it equal to one. If it has a radius of one, then the radius of the Sun is
The distance of the moon from the Earth is
We have
This is the that is the Solar radius to the lunar orbit around earth that is the molar mass of
gold to silver. This distance of the moon from the sun is about
Let us guess that the sun has to be the size that it is to be a yellow main sequence star. And that
the earth-moon system has to be 1 AU from the Sun to be in the habitable zone (to have water in
its liquid phase). Kepler’s Law of planetary motion is
Equation 2.1.
For the Sun with T the orbital period of the planet and its distance from the Sun in
astronomical units, for circular orbits. For other stars we have to include a constant k involving
the masses of the bodies:
Equation 2.2.
If the mass of the body orbiting the Star m is small compared to the mass of the star it is orbiting
we have
Equation 2.3.
The Earth is the third planet, is in the habitable zone, and its distance from the Sun defines 1
AU. Thus we ask: What is k for other star systems? For stars on the main sequence their
luminosity is proportional to their masses raised to the power of 3.5 as an estimate. We have:
Equation 2.4:
696,000k m
1,738k m
= 400
384,400k m
1,738k m
= 221.173 ≈ 220
400
220
= 1.818181...
9
5
149,592,870k m
1,738k m
= 8,608
T
2
= a
3
a
1
k
= G
M + m
4π
2
≈ G
M
4π
2
a
3
=
GM
4π
2
T
2
L =
(
M
M
⊙
)
3.5
of 15 49
Where L is in solar luminosities and M is in solar masses:
Further if we say since the Earth is in the right zone to be habitable ( ) then if a star is 100
times brighter than the Sun by the inverse square law its habitable zone is is 10
times further from the star it orbits than the Earth is from the Sun. We have
Equation 2.5:
Combining equations 2.4 and 2.5:
Equation 2.6:
For another star system we can write equation 2.3
Equation 2.7.
Where n is the number of solar masses of the star and k has M in solar masses. Combining this
with equation 2.6 we have for the habitable zone of a star:
Equation 2.8.
The luminosity of the sun is:
The separation between the earth and the sun is:
The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:
L
⊙
= 3.9E 26J/s
M
⊙
= 1.98847E 30kg
ℍ
100 = 10AU
ℍ =
(
L
L
⊙
)
ℍ =
(
M
M
⊙
)
3.5
T
2
a
3
=
k
n
T
2
=
k
n
(
M
M
⊙
)
21
4
L
0
= 3.9 × 10
26
J/s
1.5 × 10
11
m
S
0
=
39 × 10
2
4π (1.5 × 10
11
)
= 1,370wat ts /m eter
2
of 16 49
That is the effective energy hitting the earth per second per square meter. This radiation is equal
to the temperature, to the fourth power by the steffan-boltzmann constant, sigma ( ), can
be called the temperature entering, the temperature entering the earth.
intercepts the earth disc, , and distributes itself over the entire earth surface, , while
30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so
But, just as the same amount of radiation that enters the system, leaves it, to have radiative
equilibrium, the atmosphere radiates back to the surface so that the radiation from the
atmosphere, plus the radiation entering the earth, is the radiation at the surface of
the earth, . However,
And we have…
a=0.3
So, for the temperature at the surface of the Earth:
Let’s convert that to degrees centigrade:
T
e
σ
T
e
S
0
π r
2
4π r
2
σ T
e
4
=
S
0
4
(1 − a)
(1 − a)S
0
(
π r
2
4π r
2
)
σ T
a
4
σ T
e
4
σ T
s
4
σ T
a
4
= σ T
e
4
σ T
s
4
= σ T
a
4
+ σ T
e
4
= 2σ T
e
4
T
s
= 2
1
4
T
e
σ T
e
4
=
S
0
4
(1 − a)
σ = 5.67 × 10
−8
S
0
= 1,370
1,370
4
(0.7) = 239.75
T
e
4
=
239.75
5.67 × 10
−8
= 4.228 × 10
9
T
e
= 255Ke lvin
T
s
= 2
1
4
T
e
= 1.189(255) = 303Kelvin
of 17 49
Degrees Centigrade = 303 - 273 = 30 degrees centigrade
And, let’s convert that to Fahrenheit:
Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit
In reality this is warmer than the average annual temperature at the surface of the earth, but in
this model, we only considered radiative heat transfer and not convective heat transfer. In other
words, there is cooling due to vaporization of water (the formation of clouds) and due to the
condensation of water vapor into rain droplets (precipitation or the formation of rain).
The point we want to make is that the Earth has to be the distance it is from the Sun, and the
Sun has to be the size that it is. We have that
Which allows the moon to perfectly eclipse the Sun. However, this means
If the solar radius and earth orbit had to have been what they are for the Earth to be habitable,
the lunar radius and solar radius can be anything’s to satisfy this equation. This is the part
where the messenger, who we hypothesize put the moon in the sky such that it would eclipse the
sun, had a criterion to choose the size of the lunar radius and the size of its orbit. How, under
this hypothesis did it or they determine their proportions? I think the best answer would be in
equation 1.13:
Which would mean it was because of the size of the proton ( ) and its mass ( ). We derived
these values and the charge of a proton from the properties of space and time.
We have to explain the meaning of the earth day in the equation, we might be able to do this by
breaking the whole equation down into parts.
If is the angular velocity of the earth, is the mass of the moon, is the mass of the earth,
is the orbital velocity of the moon, is the orbital velocity of the earth, then we have
There are 86,400 seconds in an earth day so, the angular velocity of the earth is
SolarRa diu s
Lu n arRa diu s
= 400 =
Ear thOrbit
Lu n arOrbit
= 389
400 ≈ 389
Lu n arRa diu s
Lu n arOrbit
=
SolarRa diu s
Ear thOrbit
k v
e
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
=
1
α
4
m
2
p
⋅
h 4π r
2
p
Gc
r
p
m
p
ω
e
M
m
M
e
v
m
v
e
k v
e
⋅
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(
2π
ω
e
)
2
=
1
α
4
m
2
p
⋅
h 4π r
2
p
Gc
of 18 49
Since we have
There are 31,557,600 seconds in a year, so our value above is:
Since there are 365.25 days in a year that is
It is 6 days, which is six rotations of the earth. The whole reason then, for the hypothetical
messenger’s equation, could be to maintain our six-fold symmetry.
ω
e
=
2π
86,400s
= 7.2722E − 5
ra d
s
ω
2
e
= 5.2885E − 9
ra d
2
s
2
k v
e
= 38.5
(38.5)(4π
2
ra d
2
) = 1,520ra d
2
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(
1,520ra d
2
ω
2
e
)
=
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(2.874E11s
2
)
M
m
M
e
⋅
v
2
m
v
2
e
(536,097s) =
1
α
2
m
p
h 4π r
2
p
Gc
536,097s
31,557,600s /yr
= 0.017years
(0.017)(365.25d a ys) = 6d a ys
of 19 49
3.0 Logos
We see if we suggest there is a language of Nature that describes things universally (logos) then
to do this for planetary systems we need to work in units where their physical parameters are all
one, like earth orbit is one astronomical unit (AU). We find if we do this we get orbital velocity is
always 6 for all planets. This is nice because we have set out from the beginning guessing the
basis of Nature is in the six-fold.
With the information we have for the earth-moon-sun orbital parameters we see we can make
the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree) is the distance the earth moves
around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are
360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).
We have:
1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always
close to that because its orbit is approximately circular. We have
This is approximately the diameter of the Earth orbit. We define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
orbit rota t ion orbit m oon
29.786k m minute 1d a y k ilom eter
secon d 27.83k m d egree secon d
=
min ⋅ d ay
deg
⋅
k m
s
2
(min)(d ay) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
min ⋅ d ay
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
min
=
27.83k m
min
⋅
min
60sec
= 0.4638
k m
sec
θ
e
=
1d a y
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
of 20 49
Moon orbits:
Earth completes a 360 degree orbit yields:
Where on the right it is in radians and is the radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
Where is the inverse of the golden ratio. .
We have:
Equation 3.1
Our base ten counting is defined
is defined
, such that
which is given by
Thus since the diameter of the Earth orbit is
Then its radius is
Equation 3.2
Since we have established a connection between the microcosmos and the macrocosmos we
would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want
to know the universal gravitational constant in these units:
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786k m /s)(14,400s /deg)(1k m /s)
(0.4638k m /s)(149,598,000k m /AU )
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
of 21 49
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
Equation 3.3
Which evaluates:
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
5
3
GM
ϕ
v =
5
3
(40)(1)
0.618
= 6
AU
year
of 22 49
Converting this to meters per second:
Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s
is an average over a varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
For the earth, we show now it is true for Venus as well. Its orbital distance VU (Venus units) is
1.082E11meters. It orbital period is 1.94E7s is the Venus year (VY).
=
Thus we have
Equation 3.4
We can then express all orbital velocities as 6, but to find their values in a formal system of
units, we need to convert from these natural units to something like kg/m/s. Thus logos
translated into a language we understand can best be done in a square array, or as a matrix
transformation. We have
Equation 3.5
Where is for the moon. You will find this gives , ,
All of which are correct within the variations of these velocities in their deviations from a
perfectly circular orbit. Logos is:
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
1
2
⋅
ϕ
100
360
∘
= 1AU
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E 11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
of 23 49
Equation 3.6
Equations 3.7
Basically for the Earth
And in general for any orbiting body
Equation 3.8.
Where the right side is in the orbiting bodies’ natural units, which is its orbital radius taken as 1
divided by its year taken as 1. G and M are in these units as well, so G is constant and always 40,
and M is the mass orbited taken as 1. That is we could write
Equation 3.9.
And, again is the golden ratio conjugate is 0.618. . So,…
Equation 3.10.
Equation 3.11.
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v
e
=
5
3
GM
ϕ
= 6
AU
year
v
o
=
5
3
GM
ϕ
= 6
v
o
=
5
3
40
ϕ
= 6
ϕ
5/3 ≈ 3/4
v
o
=
3
4
GM
ϕ
= 6
v
o
=
3
4
40
ϕ
= 6
of 24 49
We want to look at the following two equations
Firstly the reason the orbital velocity of any body for circular orbits in their natural units, is 6
because 6 AU per year is approximately and is the circumference of its orbit if its orbital
radius is 1 AU, where AU is taken to be the orbital radius of the body. We also have that the
rotation of the orbiting body about its own axis if it is the Earth, has sixfold symmetry with
respect to the very atoms it is made in that we can write:
Six rotations of the Earth is given by
Equation 3.12.
That one revolution around the sun is 360 degrees and is broken up into 6 rotations of the
Earth, and the Earth year 365.25 days (rotations of earth for one revolution) is approximately 60
degrees
And, 60 degrees are the angles in an equilateral triangle, the most fundamental straight-line
form. From the last two of the above three equation we can write:
Equation 3.13.
The units on both sides are . We have taken Au/Ag to be approximately 16/9. I put the Au
(gold) and Ag (silver) because this is an equation of the Moon and the Sun, and archeologically
gold and silver have played ceremonial role in history with the Moon and the Sun, and the Sun is
gold in color and the moon is silver in color. See the Brainstorm on the next page…
k v
e
(
K E
2
moon
K E
2
earth
Ear th Da y
2
)
=
1
α
4
m
2
p
⋅
h 4π r
2
p
Gc
v
e
=
3
4
GM
ϕ
= 6
M
m
M
e
⋅
v
2
m
v
2
e
(6Da ys) =
1
α
2
m
p
h 4π r
2
p
Gc
2π
2π
1
α
2
m
p
h 4π r
2
p
Gc
M
e
M
m
⋅
v
2
e
v
2
m
= 6d ays
360degr ees
6
= 60d egrees
ϕ
Au
Ag
⋅
M
m
M
e
⋅
v
2
m
M
⊙
⋅ 6d a ys =
1
α
2
m
p
⋅ 4π r
2
p
⋅
hG
c
m
3
s ⋅ kg
of 25 49
of 26 49
4.0 Six-Fold Symmetry and The Periodic Table
Just what do we mean by six-fold unfolding? I like to start here, with a cube. It has six faces and
as such we see that six points describe a location in that putting six points at the center of each
face of the cube and connecting the points that oppose one another describes the location at its
center (O):
We let this represent
We notice each of these points on a face require two numbers in each face to describe them. So
twelve numbers become necessary:
We let this represent . This is a big reason that 6-fold
symmetry describes so much so well because 12 is the smallest
abundant number. It is evenly divisible by 1, 2, 3, 4, 6, and
1+2+3+4+6=16 is greater than 12.
Or, you can describe the point with three numbers but since there are six faces there are six
choices to do this:
We let this represent
6 ⋅ 1 = 6
2 ⋅ 6 = 12
3 ⋅ 6 = 18
of 27 49
From this we get our special ratios that are at the basis of the Natural systems which are
respectively in the the lengths of the unit lines in a plane and in space:
Which are irrational. We see them here:
These are:
But where is the golden ratio? It is in the regular pentagon:
Where is the unit 1? It is in the equilateral triangle, the most basic and stable structure, the first
2D object enclosed by straight lines:
1
2
+ 1
2
= 2
1
2
+ 2
2
= 3
2cos(π /4) = 2cos(45
∘
) = 2
2cos(π /6) = 2cos(60
∘
) = 3
2cos(π /5) = 2cos(36
∘
) =
5 + 1
2
2cos(π /3) = 2cos(30
∘
) = 1
of 28 49
of 29 49
Space is characterized by a 3 x 3 matrix because we can use it to do the cross product and form a
normal to two vectors in a plane. Let us pull out of the periodic table a 3 X 3 matrix from one we
will talk about shortly that we will see is pivotal to it.
A cube has six such faces, and each face finds it center (O) with two lines:
The center of a cube is found by 3 lines: one over, one in, one up:
3 × 3 = ar ea = 9
2 × 3 = 6
of 30 49
The periodic table is periodic over 18 groups. 2, 3 are the
smallest primes, the lowest factors in which we can factor
numbers, we have in conclusion
As such I think 6-fold symmetry is the most dynamic way in which to describe natural systems.
The atomic number Z is the number of protons characterizing an element, here we see going
vertically down the groups, it adds 8 protons to get the second line from the first, and 18 to get
the third line from the second:
That is 13-5=8 and 31-13=18, 14-6=8 and 32-14=18, and 15-7=8 and 33-15=18. Reading across is
the periods. Since period 2 and period 3 don’t have elements from groups 3 to 12, but period 3
does, we have a jump from 8 to 18, thus it would seem we can’t make one arithmetic set of
operations using addition, subtraction, multiplication, and division that describes the vertical
trend as an operation on integers that produces integers. For example:
But
Is two shy of 18. But if we try
We have
2 ⋅ 3 = 6
3 ⋅ 3 = 9
2 ⋅ 9 = 18
3 ⋅ 6 = 18
(
B C N
Al Si P
Ga Ge A s
)
⟹
5 6 7
13 14 15
31 32 33
8 ↓ 18
2
n
= 2
3
= 8
2
n
= 2
4
= 16
2
2
+ n − 2
2
3
+ 3 − 2 = 9
of 31 49
Is one to many and
Works for the other one. We can use the smallest primes 2 and 3 and the arbitrary definition
:
And we get
But then it is one shy for the 18:
But this is it seems to me is the answer, it does what is outside of these operations on integers
because it keeps these systems from interfering with one another in their interactions. It would
seem the same thing happens with the distribution of the planets around the sun. We can see
this in the Titius-Bode Law that predicts the orbits of the planets approximately which is:
We see it fails at Neptune, but the important point to make is that it only works past Mars if you
include the asteroids at n=3, and the asteroids are not a planet but a collection of rocks and
debris that couldn’t form because the other planetary orbits act on them periodically according
to Keplers Law of Planetary Motion , tearing any formation apart. If we omit the
asteroids Jupiter takes on n=4 and is then predicted to be at 2.77AU when it is at 5.2 AU. So this
is in line with our idea that protons in the elements increase in such a way through the groups,
which are defined by their common characteristics, in such a way that they don’t interfere with
one another in their necessities.
2
4
+ 4 − 2 = 18
x
0
= 1
3 ⋅ 2
2(n−1)
+ 5
3 ⋅ 2
2(1−1)
+ 5 = 8
3 ⋅ 2
(2−1)
+ 5 = 17
n = − ∞,0,1,2,...
a = 0.4 + 0.3 × 2
n
a = 0.4 + 0.3 × 2
−∞
= 0.4 = Mercur y = 0.39AU
a = 0.4 + 0.3 × 2
0
= 0.7 = Venu s = 0.72AU
a = 0.4 + 0.3 × 2
1
= 1.00 = Ear th = 1.00AU
a = 0.4 + 0.3 × 2
2
= 1.6 = Mars = 1.52AU
a = 0.4 + 0.3 × 2
3
= 2.77 = A ster oi d s
a = 0.4 + 0.3 × 2
4
= 5.2 = Jupiter = 5.20AU
a = 0.4 + 0.3 × 2
5
= 10.0 = Sat ur n = 9.58AU
a = 0.4 + 0.3 × 2
6
= 19.6 = Ura nus = 19.22AU
a = 0.4 + 0.3 × 2
7
= 38.8 ≠ Nept u ne = 30.07AU
T
2
= a
3
of 32 49
5.0 Rigorous Formulation of Proton-Seconds
We can actually formulate this differently than we have. We had
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are
lower and upper limits in an integral, then we have:
Equation 5.1
This Equation is the generalized equation we can use for solving problems. Essentially we can
rigorously formulate the notion of proton-seconds by considering
Equation 5.2
Is protons-seconds squared where current density is and ( can also be
). We say
Equation 5.3
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is
Equation 5.4
Dividing Equation 5.2 through by t:
Equation 5.5
Which is proton-seconds. Dividing through by t again:
1
t
1
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρd V
ℕ =
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
of 33 49
Equation 5.6
We see that if where and then J is I/m2 (current per square meter)
is analogous to amperes per per square meter which are coulombs per second through a surface.
Thus we are looking at a number of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:
Equation 5.7
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.
!
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h 4π r
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y
of 34 49
We are now equipt to do computations in proton-seconds. We use equation 3.6 from two to
three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal
proton that we found described its radius (My feeling is we introduce the factor of 2 because
carbon is 6 protons +6 neutrons and 2 times 6 is twelve):
Now we integrate from phosphorus is 15 protons=0.396 seconds to aluminum is 13 protons =
0.462 seconds which is to integrate across silicon (divide your answer 10 by 2 to get protons):
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 5.8
Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus
enabling the construction of logic circuits that you can use to make computing machines. But
this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the
same group as aluminum, just above it. This results in a theory for AI elements as mathematical
constructs, that we will go into now.
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.376 − 4.685) = 5protons /secon d = bor on
of 35 49
6.0 Theory of Compounds
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are
carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to
show not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 36 49
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them. But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 6.1
A = (Al, Si, P)
B = (G a, Ge, A s)
A ×
B =
B
C
N
Al Si P
Ga Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ Ga)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g /m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g /m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsin θ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(As − G a) + G e(P − Al )
SiGe
=
2B
Ge + Si
of 37 49
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 6.2
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 6.3
Equation 6.4
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − Ga) +
Ge
B
(P − Al ) =
2SiG e
Si + Ge
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x dy ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
of 38 49
If in the equation (The accurate harmonic mean form):
Equation 6.5
We make the approximation
Equation 6.6
Then the Stokes form of the equation becomes
Equation 6.7
Thus we see for this approximation there are two integrals as well:
Equation 6.8
Equation 6.9
By making the approximation
In
We have
Equation 6.10
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dydz =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )dydz =
2
3
∫
Ge
Si
dz
2SiGe
Si + Ge
≈ Ge − Si
Si(As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
of 39 49
is the differential across Si, is the differential across Ge
and is the vertical differential.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 6.11
We can go straight down group 14 to form the rest of our differentials:
It is amazing how accurately we can fit these differentials with an exponential equation
for the upward increase. The equation is
This is the halfwave:
Since B/Al=10.81/26.98=
0.40
And Ag/Cu =107.87/63.55=
1.697
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = As − Ga = 74.92 − 69.72 = 5.2
ΔSn = Sb − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 40 49
Where B and Al are the
differential across silicon
and Ag and Au are the finest
conductors of electricity, we
have
Equation 6.12
y(x) = e
(B/Al)x
+
Ag
Cu
of 41 49
Appendix 1
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
of 42 49
Appendix 2
Where,…
, ,
Radius of hydrogen atom
Remember our constant k (Don’t forget to divide by two somewhere for hydrogen packing):
Since we have the equation of the radius of a proton is given by, by evaluating it at one second
which is carbon:
And 1 second in terms of the atom is given by
Then the equation for the radius of a proton is:
Let’s verify our equation:
v =
r
t
t = 6s =
1
α
2
m
p
h 4π r
2
p
Gc
1
t
= α
2
m
p
Gc
h 4π r
2
p
v =
r
p
t
= α
2
m
p
Gc
h 4π
R
H
= 1.2E − 10m
t =
R
h
v
= R
h
⋅
1
α
2
m
p
h 4π
Gc
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
9
8
2
1
1.67E − 27
(6.626E − 34)(299,792, 459)
(6.674E − 11)4π
3
⋅
1.2E − 10
6.02E 23
= 0.93f m
of 43 49
Making the approximation 9/8~1 we can write (We have picked up the fraction 9/8 by making
several approximations)
Which gives
We form constants:
And we have the Equation:
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Thus we have
Let us see if this makes sense.
is the circumference of a unit circle and is the area of a unit circle. For the constant of
gravitation
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
2π
3
= (2π)(π)(π)
2π
π
of 44 49
is a measure of the force of space pushing back on the proton per kilogram in our theory.
This is what G measures but over square meter per kilogram . Planck’s constant h is
The speed of light is m/s thus we have
Is a measure of force over surface or,…
Is a measure of energy over a length. Thus we see we have force canceling with force and area
with area to give us the radius of a proton:
We see we have force cancelling with force, surface area cancelling with surface area ,
leaving the square kilograms (kg) of the proton. We take the square root because we are forming
a geometric mean and then we divide by the mass of a proton and multiply by the radius of a
hydrogen atom to get the proper units for the radius of a proton. That the equation makes
sense.!
G =
m
3
kg ⋅ s
2
= N
m
2
kg
2
=
N
kg
⋅
m
2
kg
N
kg
m
2
kg
h = kg
m
2
s
hc = kg
m
2
s
⋅
m
s
=
(
kg
m
s
2
)
⋅ m
2
kg
m
2
s
2
⋅ m
hc
G
=
(
kg
m
s
2
⋅ m
2
)
(
1
kg
s
2
m
⋅
kg
2
m
2
)
= kg
2
m
2
m
2
m
p
R
H
of 45 49
Appendix 3
of 46 49
of 47 49
Appendix 4
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
1
α
2
m
p
h 4π r
2
p
Gc
of 48 49
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
of 49 49
The Author
